Of the 20 lightbulbs in a box, 2 are defective. An inspector will select 2 lightbulbs simultaneously and at random from the box. What is the probability that neither of the lightbulbs selected will be defective?

Give your answer as a fraction.




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P(neither lightbulb is defective) = P(BOTH lightbulbs work)
= P(1st lightbulb works AND 2nd lightbulb works)
= P(1st lightbulb works) x P(2nd lightbulb works)
= 18/20 x 17/19
= 9/10 x 17/19 [since 18/20 = 9/10]
= 153/190

Answer: 153/190

ASIDE: P(1st lightbulb works) = 18/20, because there are 20 lightbulbs and 18 of them work.
P(2nd lightbulb works) = 17/19 because, once we have selected a working bulb for the 1st selection, there are 19 lightbulbs remaining and 17 of them work.

Cheers,
Brent
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YOu are going wrong because the answer you are getting is for a different combination..
1-P(both defective) will give you Probability when both are not defective, so it will contain ways when one is defective and one is not AND this i snot the same as the probability
that neither of the lightbulbs selected will be defective?
..
so your complement method should be \(1-P(both -not- defective)-P(just- one -is -defective)=1-\frac{2*1}{20*19}-2*\frac{2*18}{20*19}=1-\frac{1}{190}-\frac{36}{190}=1-\frac{37}{190}=\frac{153}{190}\)..
You are multiplying just one defective by two as it can be defective first and then non-defective or non-defective first and then defective.
Of course here straight method is easier
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